# NCERT Solutions for Exercise 6.4 Class 12 Maths Chapter 10 - Application of Derivatives

NCERT solutions for exercise 6.4 Class 12 Maths chapter 6 discuss problems related to the approximation of certain quantities using differentiation. Prior to exercise 6.4 Class 12 Maths, there are three exercises in which the concepts of rate of change of quantities, increasing and decreasing function and tangents and normals are discussed. Till the NCERT solutions for Class 12 Maths chapter 6 exercise 6.4 NCERT Book presents 25 solved examples. These NCERT syllabus solved examples give an insight into the topics covered in the chapter. The Class 12 Maths chapter 6 exercise 6.4 gives a detailed explanation of numerical related to the approximation of quantities. As mentioned, other than Class 12th Maths chapter 6 exercise 6.4 there are 5 exercises including the miscellaneous exercise.

• Application of Derivatives Exercise 6.1

• Application of Derivatives Exercise 6.2

• Application of Derivatives Exercise 6.3

• Application of Derivatives Exercise 6.5

• Application of Derivatives Miscellaneous Exercise

## Application of Derivatives Class 12 Chapter 6 Exercise 6.4

Question:1(i) Using differentials, find the approximate value of each of the following up to 3
places of decimal. $\sqrt {25.3 }$

Lets suppose $y = \sqrt x$ and let x = 25 and $\Delta x = 0.3$
Then,
$\Delta y = \sqrt{x+\Delta x} - \sqrt x$
$\Delta y = \sqrt{25+0.3} - \sqrt 25$
$\Delta y = \sqrt{25.3} - 5$
$\sqrt{25.3} = \Delta y +5$
Now, we can say that $\Delta y$ is approximate equals to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.3) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.3)\\ dy = \frac{1}{2\sqrt 25}.(0.3)\\ dy = \frac{1}{10}.(0.3)\\ dy = 0.03$
Now,
$\sqrt{25.3} = \Delta y +5\\ \sqrt {25.3} = 0.03 + 5\\ \sqrt{25.3} = 5.03$
Hence, $\sqrt{25.3}$ is approximately equals to 5.03

Question:1(ii) Using differentials, find the approximate value of each of the following up to 3 places of decimal.

$\sqrt { 49.5 }$

Lets suppose $y = \sqrt x$ and let x = 49 and $\Delta x = 0.5$
Then,
$\Delta y = \sqrt{x+\Delta x} - \sqrt x$
$\Delta y = \sqrt{49+0.5} - \sqrt 49$
$\Delta y = \sqrt{49.5} - 7$
$\sqrt{49.5} = \Delta y +7$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.5)\\ dy = \frac{1}{2\sqrt 49}.(0.5)\\ dy = \frac{1}{14}.(0.5)\\ dy = 0.035$
Now,
$\sqrt{49.5} = \Delta y +7\\ \sqrt {49.5} = 0.035 + 7\\ \sqrt{49.5} = 7.035$
Hence, $\sqrt{49.5}$ is approximately equal to 7.035

Question:1(iii) Using differentials, find the approximate value of each of the following up to 3 places of decimal.

$\sqrt {0.6}$

Lets suppose $y = \sqrt x$ and let x = 1 and $\Delta x = -0.4$
Then,
$\Delta y = \sqrt{x+\Delta x} - \sqrt x$
$\Delta y = \sqrt{1+(-0.4)} - \sqrt 1$
$\Delta y = \sqrt{0.6} - 1$
$\sqrt{0.6} = \Delta y +1$
Now, we cam say that $\Delta y$ is approximately equals to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(-0.4) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = -0.4)\\ dy = \frac{1}{2\sqrt 1}.(-0.4)\\ dy = \frac{1}{2}.(-0.4)\\ dy = -0.2$
Now,
$\sqrt{0.6} = \Delta y +1\\ \sqrt {0.6} = (-0.2) + 1\\ \sqrt{0.6} = 0.8$
Hence, $\sqrt{0.6}$ is approximately equal to 0.8

Question:1(iv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$( 0.009 ) ^{1/3 }$

Lets suppose $y = (x)^{\frac{1}{3}}$ and let x = 0.008 and $\Delta x = 0.001$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}$
$\Delta y = ({0.008+ 0.001})^{\frac{1}{3}} - (0.008)^{\frac{1}{3}}$
$\Delta y = ({0.009})^{\frac{1}{3}} - 0.2$
$({0.009})^{\frac{1}{3}} = \Delta y + 0.2$
Now, we cam say that $\Delta y$ is approximately equals to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(0.001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = 0.001)\\ dy = \frac{1}{3(0.008)^{\frac{2}{3}}}.(0.001)\\ dy = \frac{1}{0.12}.(0.001)\\ dy = 0.008$
Now,
$(0.009)^{\frac{1}{3}} = \Delta y +0.2\\ (0.009)^{\frac{1}{3}} = (0.008) + 0.2\\ (0.009)^{\frac{1}{3}} = 0.208$
Hence, $(0.009)^{\frac{1}{3}}$ is approximately equal to 0.208

Question:1(v) Using differentials, find the approximate value of each of the following up to 3
places of decimal.

$( 0.999) ^{1/10 }$

Lets suppose $y = (x)^{\frac{1}{10}}$ and let x = 1 and $\Delta x = -0.001$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{10}} - (x)^{\frac{1}{10}}$
$\Delta y = ({1 - 0.001})^{\frac{1}{10}} - (1)^{\frac{1}{10}}$
$\Delta y = ({0.999})^{\frac{1}{10}} - 1$
$({0.999})^{\frac{1}{10}} = \Delta y + 1$
Now, we cam say that $\Delta y$ is approximately equals to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{10 (x)^{\frac{9}{10}}}.(-0.001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{10}} \ and \ \Delta x = -0.001)\\ dy = \frac{1}{10(1)^{\frac{9}{10}}}.(-0.001)\\ dy = \frac{1}{10}.(-0.001)\\ dy = -0.0001$
Now,
$(0.999)^{\frac{1}{10}} = \Delta y +1\\ (0.999)^{\frac{1}{10}} = (-0.0001) + 1\\ (0.999)^{\frac{1}{10}} = 0.9999 = 0.999 \ upto \ three\ decimal \ place$
Hence, $(0.999)^{\frac{1}{10}}$ is approximately equal to 0.999 (because we need to answer up to three decimal place)

Question:1(vi) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$(15 )^{1/4}$

Let's suppose $y = (x)^{\frac{1}{4}}$ and let x = 16 and $\Delta x = -1$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}$
$\Delta y = ({16 - 1})^{\frac{1}{4}} - (16)^{\frac{1}{4}}$
$\Delta y = ({15})^{\frac{1}{4}} - 2$
$({15})^{\frac{1}{4}} = \Delta y + 2$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = -1)\\ dy = \frac{1}{4(16)^{\frac{3}{4}}}.(-1)\\ dy = \frac{1}{4\times 8}.(-1)\\dy = \frac{1}{32}.(-1) \\dy = -0.031$
Now,
$(15)^{\frac{1}{4}} = \Delta y +2\\ (15)^{\frac{1}{4}} = (-0.031) + 2\\ (15)^{\frac{1}{4}} = 1.969$
Hence, $(15)^{\frac{1}{4}}$ is approximately equal to 1.969

Question:1(vii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$(26)^{1/3 }$

Lets suppose $y = (x)^{\frac{1}{3}}$ and let x = 27 and $\Delta x = -1$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}$
$\Delta y = ({27 - 1})^{\frac{1}{3}} - (27)^{\frac{1}{3}}$
$\Delta y = ({26})^{\frac{1}{3}} - 3$
$({26})^{\frac{1}{3}} = \Delta y + 3$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = -1)\\ dy = \frac{1}{3(27)^{\frac{2}{3}}}.(-1)\\ dy = \frac{1}{3\times 9}.(-1)\\dy = \frac{1}{27}.(-1) \\dy = -0.037$
Now,
$(27)^{\frac{1}{3}} = \Delta y +3\\ (27)^{\frac{1}{3}} = (-0.037) + 3\\ (27)^{\frac{1}{3}} = 2.963$
Hence, $(27)^{\frac{1}{3}}$ is approximately equal to 2.963

Question:1(viii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$( 255) ^{1/4}$

Let's suppose $y = (x)^{\frac{1}{4}}$ and let x = 256 and $\Delta x = -1$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}$
$\Delta y = ({256 - 1})^{\frac{1}{4}} - (256)^{\frac{1}{4}}$
$\Delta y = ({255})^{\frac{1}{4}} - 4$
$({255})^{\frac{1}{4}} = \Delta y + 4$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = -1)\\ dy = \frac{1}{4(256)^{\frac{3}{4}}}.(-1)\\ dy = \frac{1}{4\times 64}.(-1)\\dy = \frac{1}{256}.(-1) \\dy = -0.003$
Now,
$(255)^{\frac{1}{4}} = \Delta y +4\\ (255)^{\frac{1}{4}} = (-0.003) + 4\\ (255)^{\frac{1}{4}} = 3.997$
Hence, $(255)^{\frac{1}{4}}$ is approximately equal to 3.997

Question:1(ix) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$( 82) ^{1/4 }$

Let's suppose $y = (x)^{\frac{1}{4}}$ and let x = 81 and $\Delta x = 1$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}$
$\Delta y = ({81 + 1})^{\frac{1}{4}} - (81)^{\frac{1}{4}}$
$\Delta y = ({82})^{\frac{1}{4}} - 3$
$({82})^{\frac{1}{4}} = \Delta y + 3$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = 1)\\ dy = \frac{1}{4(81)^{\frac{3}{4}}}.(1)\\ dy = \frac{1}{4\times 27}.(1)\\dy = \frac{1}{108}.(1) \\dy = .009$
Now,
$(82)^{\frac{1}{4}} = \Delta y +3\\ (82^{\frac{1}{4}} = (0.009) + 3\\ (82)^{\frac{1}{4}} = 3.009$
Hence, $(82)^{\frac{1}{4}}$ is approximately equal to 3.009

Question:1(x) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$( 401 ) ^{1/2 }$

Let's suppose $y = (x)^{\frac{1}{2}}$ and let x = 400 and $\Delta x = 1$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}$
$\Delta y = ({400 + 1})^{\frac{1}{2}} - (400)^{\frac{1}{2}}$
$\Delta y = ({401})^{\frac{1}{2}} - 20$
$({401})^{\frac{1}{2}} = \Delta y + 20$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 1)\\ dy = \frac{1}{2(400)^{\frac{1}{2}}}.(1)\\ dy = \frac{1}{2\times 20}.(1)\\dy = \frac{1}{40}.(1) \\dy = 0.025$
Now,
$(401)^{\frac{1}{2}} = \Delta y +20\\ (401)^{\frac{1}{2}} = (0.025) + 20\\ (401)^{\frac{1}{2}} = 20.025$
Hence, $(401)^{\frac{1}{2}}$ is approximately equal to 20.025

Question:1(xi) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$( 0.0037 ) ^{1/2 }$

Lets suppose $y = (x)^{\frac{1}{2}}$ and let x = 0.0036 and $\Delta x = 0.0001$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}$
$\Delta y = ({0.0036 + 0.0001})^{\frac{1}{2}} - (0.0036)^{\frac{1}{2}}$
$\Delta y = ({0.0037})^{\frac{1}{2}} - 0.06$
$({0.0037})^{\frac{1}{2}} = \Delta y + 0.06$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(0.0001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 0.0001)\\ dy = \frac{1}{2(0.0036)^{\frac{1}{2}}}.(0.0001)\\ dy = \frac{1}{2\times 0..06}.(0.0001)\\dy = \frac{1}{0.12}.(0.0001) \\dy = 0.0008$
Now,
$(0.0037)^{\frac{1}{2}} = \Delta y +0.06\\ (0.0037)^{\frac{1}{2}} = (0.0008) + 0.06\\ (0.0037)^{\frac{1}{2}} = 0.0608$
Hence, $(0.0037)^{\frac{1}{2}}$ is approximately equal to 0.060 (because we need to take up to three decimal places)

Question:1(xii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$(26.57) ^ {1/3}$

Lets suppose $y = (x)^{\frac{1}{3}}$ and let x = 27 and $\Delta x = -0.43$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{3}} - (x)^{\frac{1}{3}}$
$\Delta y = ({27 - 0.43})^{\frac{1}{3}} - (27)^{\frac{1}{3}}$
$\Delta y = ({26.57})^{\frac{1}{3}} - 3$
$({26.57})^{\frac{1}{3}} = \Delta y + 3$
Now, we cam say that $\Delta y$ is approximately equals to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{3 (x)^{\frac{2}{3}}}.(-0.43) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{3}} \ and \ \Delta x = -0.43)\\ dy = \frac{1}{3(27)^{\frac{2}{3}}}.(-0.43)\\ dy = \frac{1}{3\times 9}.(-0.43)\\dy = \frac{1}{27}.(-0.43) \\dy = -0.0159 = -0.016 (approx.)$
Now,
$(26.57)^{\frac{1}{3}} = \Delta y +3\\ (26.57)^{\frac{1}{3}} = (-0.016) + 3\\ (26.57)^{\frac{1}{3}} = 2.984$
Hence, $(0.0037)^{\frac{1}{2}}$ is approximately equal to 0.060 (because we need to take up to three decimal places)

Question:1(xiii) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$( 81.5 ) ^{1/4 }$

Lets suppose $y = (x)^{\frac{1}{4}}$ and let x = 81 and 0.5
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}$
$\Delta y = ({81 + 0.5})^{\frac{1}{4}} - (81)^{\frac{1}{4}}$
$\Delta y = ({81.5})^{\frac{1}{4}} - 3$
$({81.5})^{\frac{1}{4}} = \Delta y + 3$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = 0.5)\\ dy = \frac{1}{4(81)^{\frac{3}{4}}}.(0.5)\\ dy = \frac{1}{4\times 27}.(0.5)\\dy = \frac{1}{108}.(0.5) \\dy = .004$
Now,
$(81.5)^{\frac{1}{4}} = \Delta y +3\\ (82^{\frac{1}{4}} = (0.004) + 3\\ (82)^{\frac{1}{4}} = 3.004$
Hence, $(81.5)^{\frac{1}{4}}$ is approximately equal to 3.004

Question:1(xiv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.

$( 3.968) ^{3/2 }$

Let's suppose $y = (x)^{\frac{3}{2}}$ and let x = 4 and $\Delta x = -0.032$
Then,
$\Delta y = ({x+\Delta x})^{\frac{3}{2}} - (x)^{\frac{3}{2}}$
$\Delta y = ({4 - 0.032})^{\frac{3}{2}} - (4)^{\frac{3}{2}}$
$\Delta y = ({3.968})^{\frac{3}{2}} - 8$
$({3.968})^{\frac{3}{2}} = \Delta y + 8$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{3 (x)^{\frac{1}{2}}}{2}.(-0.032) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{3}{2}} \ and \ \Delta x = -0.032)\\ dy = \frac{3 (4)^{\frac{1}{2}}}{2}.(-0.032)\\ dy = \frac{3\times 2}{2}.(-0.032)\\\\dy = -0.096$
Now,
$(3.968)^{\frac{3}{2}} = \Delta y +8\\ (3.968)^{\frac{3}{2}} = (-0.096) + 8\\ (3.968)^{\frac{3}{2}} = 7.904$
Hence, $(3.968)^{\frac{3}{2}}$ is approximately equal to 7.904

Question:1(xv) Using differentials, find the approximate value of each of the following up to 3
places of decimal.
$( 32.15 ) ^{1/5}$

Lets suppose $y = (x)^{\frac{1}{5}}$ and let x = 32 and $\Delta x = 0.15$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{5}} - (x)^{\frac{1}{5}}$
$\Delta y = ({32 + 0.15})^{\frac{1}{5}} - (32)^{\frac{1}{5}}$
$\Delta y = ({32.15})^{\frac{1}{5}} - 2$
$({32.15})^{\frac{1}{5}} = \Delta y + 2$
Now, we can say that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{5 (x)^{\frac{4}{5}}}.(0.15) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{5}} \ and \ \Delta x = 0.15)\\ dy = \frac{1 }{5 (32)^{\frac{4}{5}}}.(0.15)\\ dy = \frac{1}{5\times16}.(0.15)\\\\dy = \frac{0.15}{80}\\ dy = 0.001$
Now,
$(32.15)^{\frac{1}{5}} = \Delta y +2\\ (32.15)^{\frac{1}{5}} = (0.001) + 2\\ (32.15)^{\frac{1}{5}} = 2.001$
Hence, $(32.15)^{\frac{1}{5}}$ is approximately equal to 2.001

Question:2 Find the approximate value of f (2.01), where $f (x) = 4x^2 + 5x + 2.$

Let x = 2 and $\Delta x = 0.01$
$f(x+\Delta x) = 4(x+\Delta x)^2 +5(x+\Delta x)+2$
$\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)$
We know that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}.\Delta x\\ dy = (8x+5).(0.01) \ \ \ \ \ \ \ \ \ (\because y = f(x) = 4x^2+5x+2 \ and \ \Delta x = 0.01)\\ dy = 0.08x+0.05$
$f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.08x + 0.05 + 4x^2 + 5x +2\\ f(x+\Delta x) = 0.08(2)+0.05+4(2)^2+5(2)+2\\ f(x+\Delta x) = 0.16 + 0.05 + 16 + 10 + 2\\ f(x+\Delta x) = 28.21$
Hence, the approximate value of f (2.01), where $f (x) = 4x^2 + 5x + 2.$ is 28.21

Question:3 Find the approximate value of f (5.001), where $f (x) = x^3 - 7x^2 + 15.$

Let x = 5 and $\Delta x = 0.001$
$f(x+\Delta x) =(x+\Delta x)^3 - 7(x+\Delta x)^2 +15$
$\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)$
We know that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}.\Delta x\\ dy = (3x^2 - 14x).(0.001) \ \ \ \ \ \ \ \ \ (\because y = f(x) = x^3-7x^2+15 \ and \ \Delta x = 0.001)\\ dy =0.003x^2 -0.014x$
$f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.003x^2 - 0.014x + x^3 - 7x^2 +15\\ f(x+\Delta x) =0.003(5)^2-0.014(5)+(5)^3-7(5)^2+15\\ f(x+\Delta x) = 0.075-0.07+125-175+15\\ f(x+\Delta x) = -34.995$
Hence, the approximate value of f (5.001), where $f (x) = x^3 - 7x^2 + 15\ is \ -34.995$

Question:4 Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.

Side of cube increased by 1% = 0.01x m
Volume of cube = $x^3 \ m^3$
we know that $\Delta y$ is approximately equal to dy
So,
$dy = \frac{dy}{dx}.\Delta x\\ dy =3x^2(0.01x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = x^3 \ and \ \Delta x = 0.01x)\\ dy = 0.03x^3$
Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 1% is $0.03x^3 \ m^3$

Question:5 Find the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%.

Side of cube decreased by 1% $(\Delta x)$ = -0.01x m
The surface area of cube = $6a^2 \ m^2$
We know that, $(\Delta y)$ is approximately equal to dy

$dy = \frac{dy}{dx}.\Delta x\\ dy = 12a(-0.01x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = 6a^2 \ and \ \Delta x = -0.01x)\\ dy = 12x(-0.01x)\\ dy=-0.12x^2 \ m^2$
Hence, the approximate change in the surface area of a cube of side x metres
caused by decreasing the side by 1%. is $-0.12x^2 \ m^2$

Question:6 If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

Error in radius of sphere $(\Delta r)$ = 0.02 m
Volume of sphere = $\frac{4}{3}\pi r^3$
Error in volume $(\Delta V)$
$dV = \frac{dV}{dr}.\Delta r\\ dV = 4\pi r^2 .\Delta r \ \ \ \ \ \ \ \ \ \ \ \ \ (\because V = \frac{4}{3}\pi r^3, r =7 \ and \ \Delta r = 0.02 )\\ dV = 4\pi (7)^2 (0.02)\\ dV= 4\pi (49) (0.02)\\ dV = 3.92\pi$
Hence, the approximate error in its volume is $3.92\pi \ m^3$

Question:7 If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.

Error in radius of sphere $(\Delta r)$ = 0.03 m
The surface area of sphere = $4\pi r^2$
Error in surface area $(\Delta A)$
$dA = \frac{dA}{dr}.\Delta r\\ dA = 8\pi r .\Delta r \ \ \ \ \ \ \ \ \ \ \ \ \ (\because A = 4\pi r^2, r =9 \ and \ \Delta r = 0.03 )\\ dA = 8\pi (9) (0.03)\\ dA= 2.16\pi$
Hence, the approximate error in its surface area is $2.16\pi \ m^2$

Question:8 If $f(x) = 3x ^2 + 15x + 5$ , then the approximate value of f (3.02) is
(A) 47.66 (B) 57.66 (C) 67.66 (D) 77.66

Let x = 3 and $\Delta x = 0.02$
$f(x+\Delta x) = 3(x+\Delta x)^2 +15(x+\Delta x)+5$
$\Delta y = f(x+\Delta x) - f(x)\\ f(x+\Delta x) = \Delta y + f(x)$
We know that $\Delta y$ is approximately equal to dy
$dy = \frac{dy}{dx}.\Delta x\\ dy = (6x+15).(0.02) \ \ \ \ \ \ \ \ \ (\because y = f(x) = 3x^2+15x+5 \ and \ \Delta x = 0.02)\\ dy = 0.12x+0.3$
$f(x+\Delta x) = \Delta y + f(x)\\ f(x+\Delta x) = 0.12x + 0.3 + 3x^2 + 15x +5\\ f(x+\Delta x) = 0.12(3)+0.3+3(3)^2+15(3)+5\\ f(x+\Delta x) = 0.36+ 0.3 + 27 + 45 + 5\\ f(x+\Delta x) = 77.66$
Hence, the approximate value of f (3.02) is 77.66
Hence, (D) is the correct answer

Question:9 The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
(A) 0.06 $x^3 \ m^3$ (B) 0.6 $x^3 \ m^3$ (C) 0.09 $x^3 \ m^3$ (D) 0.9 $x^3 \ m^3$

Side of cube increased by 3% = 0.03x m
The volume of cube = $x^3 \ m^3$
we know that $\Delta y$ is approximately equal to dy
So,
$dy = \frac{dy}{dx}.\Delta x\\ dy =3x^2(0.03x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = x^3 \ and \ \Delta x = 0.03x)\\ dy = 0.09x^3$
Hence, the approximate change in volume V of a cube of side x metres caused by increasing the side by 3% is $0.09x^3 \ m^3$
Hence, (C) is the correct answer

## More About NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4

Nine questions and their explanations are given in exercise 6.4 Class 12 Maths solutions. There are two objectives questions in the NCERT solutions for Class 12 Maths chapter 6 exercise 6.4. The first question of Class 12 Maths chapter 6 exercise 6.4 have 15 sub questions. All these questions are detailed in the Class 12th Maths chapter 6 exercise 6.4

Also Read| Application of Derivatives Class 12 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.4

• Exercise 6.4 Class 12 Maths gives more clarity about the approximation technique using differentiation for certain quantities

• One question from Class 12 Maths chapter 6 exercise 6.4 may appear for the Class 12 CBSE Exam.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 6

• NCERT Solutions for Class 12 Maths Chapter 6

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12th Maths

• NCERT Exemplar Class 12th Physics

• NCERT Exemplar Class 12th Chemistry

• NCERT Exemplar Class 12th Biology